Answer:
in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution. (option C).
Explanation:
In a reaction where NaOH is added to H2O.
NaOH is considered a strong base, this means that in an aqueous solution ( in water) it's able to completely disassociate in ions.
There will not remain any NaOH in the solution. This means option D is not correct.
The ions in which NaOH will disassociate are : NaOH → Na+ + OH-
These ions we will find in the solution.
Not only Na+ because NaOH is a strong base, so there will be a lot of OH- ions as well in solution.
This means in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution.
The question is incomplete, complete question is :
A chemist must dilute 73.9 mL of 400 mM aqueous sodium carbonate solution until the concentration falls to 125 mM . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.
Answer:
The final volume of the solution will be 0.236 L.
Explanation:
Concentration of sodium carbonate solution before dilution =
Volume of sodium carbonate solution before dilution = 
Concentration of sodium carbonate solution after dilution =
Volume of sodium carbonate solution after dilution = 
Dilution equation is given by:
1 mL = 0.001 L
236 mL = 0.236 L
The final volume of the solution will be 0.236 L.
The bucket with lead added would replace more water, as it would require more volume to achieve the same mass.
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
We are given that;
1 stadium holds = 1 × 10^5 people
Number of iron atoms is 1 × 10^18 atoms
Assuming the stadium would carry an equivalent number of atoms as people.
Then, 1 stadium will carry 1 × 10^5 atoms
Therefore,
To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
